Han Zhi Yuan's answer to universe's Secondary 4 A Maths Singapore question.
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y = kx^2 + (k+1)x + k.
When k = -3,
y = -3x^2 - 2x - 3.
As the coefficient of x^2 is negative, the quadratic curve is pointed downwards and the curve is decreasing after the maximum point.
By using completing the square form:
y = -3x^2 - 2x - 3
= -3[x^2 + (2/3)x + 1]
= -3[(x + 1/3)^2 + 8/9]
The maximum of the curve is when x = -1/3, y = -3 * 8/9 = -8/3.
So the range of values of x when the curve is decreasing is x > -1/3.
When k = -3,
y = -3x^2 - 2x - 3.
As the coefficient of x^2 is negative, the quadratic curve is pointed downwards and the curve is decreasing after the maximum point.
By using completing the square form:
y = -3x^2 - 2x - 3
= -3[x^2 + (2/3)x + 1]
= -3[(x + 1/3)^2 + 8/9]
The maximum of the curve is when x = -1/3, y = -3 * 8/9 = -8/3.
So the range of values of x when the curve is decreasing is x > -1/3.
Date Posted:
3 years ago
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